-5(z-5)+2=10-(z-1)

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Solution for -5(z-5)+2=10-(z-1) equation: 



-5(z-5)+2=10-(z-1)

We move all terms to the left:

-5(z-5)+2-(10-(z-1))=0

We multiply parentheses

-5z-(10-(z-1))+25+2=0



We calculate terms in parentheses: -(10-(z-1)), so:

10-(z-1)

determiningTheFunctionDomain
-(z-1)+10

We get rid of parentheses

-z+1+10

We add all the numbers together, and all the variables

-1z+11

Back to the equation:

-(-1z+11)



We add all the numbers together, and all the variables

-5z-(-1z+11)+27=0

We get rid of parentheses

-5z+1z-11+27=0

We add all the numbers together, and all the variables

-4z+16=0

We move all terms containing z to the left, all other terms to the right

-4z=-16

z=-16/-4

z=+4

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